axv:1401.6667v2 [math.nt] 31 Jan 2014 Ranks of quotents, emandes and p-adc dgts of matces Mustafa Elshekh Andy Novocn Mak Gesbecht Abstact Fo a pme p and a matx A Z n n, wte A as A = p(a quo p)+ (A em p) whee the emande and quotent opeatons ae appled element-wse. Wte the p-adc expanson of A as A = A [0] + pa [1] + p 2 A [2] + whee each A [] Z n n has entes between [0, p 1]. Uppe bounds ae poven fo the Z-anks of A em p, and A quo p. Also, uppe bounds ae poven fo the Z/pZ-ank of A [] fo all 0 when p = 2, and a conjectue s pesented fo odd pmes. Keywods: Matx ank, Intege matx, Remande and quotent, p-adc expanson. AMS classfcaton: 15A03, 15B33, 15B36, 11C20. Outlne Ths pape pesents two elated esults on ntege matces afte applyng element-wse dvson wth emande. Fst, let A be an n n ntege matx wth ank ove Z and ank 0 ove Z/pZ. If n > p 0 then Theoem 1 n Secton 1 shows that ank(a em p) (p 0 1)(p + 1)/(2(p 1)) and ank(a quo p) + (p 0 1)(p + 1)/(2(p 1)). The second esult s concened wth the Z/pZ-anks of p-adc dgts of an ntege matx. Let U, S, V Z n n such that U, V have entes fom {0, 1}, det U det V 0 (mod 2), S = dag(1,..., 1, 0,..., 0), be the ank of S ove Cheton School of Compute Scence, Unvesty of Wateloo, Wateloo, Ontao, Canada (melshek@uwateloo.ca, andy@novocn.com, mwg@uwateloo.ca). Suppoted by the Natual Scences and Engneeng Reseach Councl (NSERC) of Canada. 1
Ranks of quotents, emandes and p-adc dgts of matces 2 Z/2Z, and n 2. If M = USV Z n n, then Theoem 16 n Secton 2 shows that ank of M [] ove Z/2Z s ( ) 2 fo all 1. A conjectue s pesented n Secton 2.3 fo the same setup, but fo p an odd pme. A esult on ntege ank of Latn squaes s also obtaned. Let A be the ntege matx of ank one fomed by the oute poduct between the vecto (1, 2,..., p 1) and ts tanspose. Then A em p s a Latn squae on the symbols {1,..., p 1}. It s shown n Coollay 10 n Secton 1.3 that the ntege ank of ths Latn squae s (p + 1)/2. 1 Quotent and Remande Matces Fo any ntege n and any pme p, let n em p and n quo p denote the (nonnegatve) emande and quotent n the Eucldean dvson n = qp + whee 0 < p. The opeatos em p and quo p ae natually extended to vectos and matces usng element-wse applcaton. Thoughout, we utlze the noton of Smth nomal fom of an ntege matx. Fo any matx A Z n n of ank, thee exst unmodula matces U, V Z n n and a unque n n ntege matx S = dag(s 1, s 2,..., s n ) such that A = USV. Futhemoe, s s +1 fo all 1 n and s = 0 fo all < n. S s called the Smth nomal fom of A. Fo a dscusson on exstence and unqueness of Smth nomal fom, we efe to the eade to the textbook by Newman [3]. We use two notons of anks. The ntege ank of A Z n n s denoted by ank(a). The ank of the mage of A n the fnte feld Z/pZ s denoted by ank p (A). Altenatvely, f = ank(a) and the Smth fom of A s S = dag(s 1,..., s, 0,..., 0), then ank p (A) = 0 s the maxmal ndex such that p s. Fnally, we use the notaton A,j fo the jth column of A Z n n and a,j fo the enty (, j) of A. 1.1 Rank Theoem The followng theoem s the man esult of Secton 1. Theoem 1. Let A be an n n matx ove Z, = ank(a), 0 = ank p (A), and assume n > p 0. Then () ank(a em p) (p 0 1)(p + 1)/(2(p 1)). () ank(a quo p) + (p 0 1)(p + 1)/(2(p 1)).
Ranks of quotents, emandes and p-adc dgts of matces 3 Poof. We wll pove pat () n Lemma 2. Fo pat (), we have A = (A em p) + p(a quo p), o p(a quo p) = A (A em p). Fo matces X = Y + Z, ank s sub-addtve and ank(x) ank(y ) + ank(z). Scalng a matx by p o 1 does not change ts ank. So ank(a quo p) ank(a) + ank(a em p) = + ank(a em p). Lemma 2. ank(a em p) (p 0 1)(p + 1)/(2(p 1)). Poof. Let A = USV be the Smth nomal fom of A, wth S = S + ps q whee S q = S quo p and S = S em p. Then A em p = USV em p = (US V + pus q V ) em p = US V em p. (1) If 0 = ank p (A) then S = dag(σ 1,..., σ 0, 0,..., 0) whee σ [1, p 1] fo all 1 0. The jth column of A em p s ( 0 ) ( 0 ) A,j em p = σ l v l,j U,l em p = c l,j U,l em p, (2) l=1 whee c l,j [0, p 1]. If we only consde the non-zeo coeffcents c l,j, then the ght-hand sde of (2) s an -tem sum (c l1,ju,l1 +... + c l,ju,l ) em p, whee 1 0 and 1 l 1 < l 2 <... < l 0. The coeffcents c lk,j ae elements n [1, p 1] whch ae unts modulo p. In patcula, we can facto c l1,j fom the sum, and e-wte (2) as: A,j em p = (c l1,j(u,l1 + α l2,ju l2,j +... + α l,ju,l )) em p, (3) whee α lk,j [1, p 1] fo all k. Fx some, j and some non-zeo assgnment of α l2,j,..., α l,j n (3) and let û = U,l1 + α l2,ju l2,j... + α l,ju,l. Then (3) becomes A,j em p = (c l1,jû) em p. Thee ae p 1 possble values fo c l1,j and hence the possble values of A,j em p ae: {û em p, (2û) em p, ((p 1)û) em p}. (4) We ae nteested n gettng an uppe bound on the ank of ths set of vectos. Fst note that (xy) em p = (x em p)(y em p) em p. So (û) em p = ((û em p)) em p fo [1, p 1]. Hence the maxmal ank one can acheve fom (4) occus when (up to pemutaton) û em p = (0, 1, 2,..., p 1,...). The est of the entes ae duplcates fom the same ange [0, p 1] by the l=1
Ranks of quotents, emandes and p-adc dgts of matces 4 pgeonhole pncple. Now apply Lemma 3 to conclude that the vectos n (4) have ank at most (p + 1)/2. Thus fo each, j and non-zeo assgnment of α l2,j,..., α l,j, thee ae at most (p + 1)/2 lnealy ndependent columns of A em p. We now count the maxmal possble numbe of dstnct A,j s. Thee ae ( 0 ) possble ways to select dffeent columns fom the fst 0 columns of U. Fo each choce, thee ae 1 coeffcents: α l2,j,..., α l,j, and (p 1) 1 possble ways to assgn the non-zeo values fom [1, p 1]. Each choce gves a set of vectos as n (4) whose ank s at most (p + 1)/2. Summng ove all [1, 0 ], the maxmal possble ank fom the span of columns n (2) s 0 ( 0 =1 usng the bnomal theoem. ) (p ) 1 p + 1 2 1.2 Remande of Rank-1 Matces In ths secton we pove the followng auxlay esult. = p 0 1 p + 1 p 1 2, (5) Lemma 3. Let p be any odd pme, n p. Let u Z n be any non-zeo vecto whee the entes of u em p nclude {1, 2,..., p 1}. Then the set of vectos {u em p, (2u) em p,..., ((p 1)u) em p} s lnealy dependent and has ank (p + 1)/2. Fst we pove ths esult fo n = p 1. A genealzaton follows. Let u = (1, 2,..., p 1) Z (p 1) and M Z (p 1) (p 1) be the ank-1 matx M = uu T and R = M em p. Lemma 4. ank(r) = (p + 1)/2. Poof. Lemma 5 shows that (p + 1)/2 s an uppe bound on the ank and Lemma 7 shows that (p + 1)/2 s a lowe bound. Lemma 5. ank(r) (p + 1)/2. Poof. Let 1 j (p 1)/2 and 1 p 1. Wte j = qp + whee 0 < p. Also, j < p = p p j, whch mples 0. Then (p j) = ( q 1)+(p ) whee 0 < (p ) < p. So j em p+(p j) em p = + (p ) = p. But R,j = j em p, so fo all 1 (p 1)/2 we have R, = (p, p,..., p) T R,p. Thus thee ae (p 1)/2 lnealy dependent columns, and no moe than (p + 1)/2 lnealy ndependent columns.
Ranks of quotents, emandes and p-adc dgts of matces 5 To pove that (p + 1)/2 s also a lowe bound on the ank, t suffces (usng Lemma 5) to consde the matx B of sze (p 1) p+1 whch s 2 fomed by the fst (p 1)/2 columns of R and the column B,(p+1)/2 = R,(p+1)/2 +R,(p 1)/2 = (p,..., p) T. The matx B has the followng stuctue: p 1 1 2 p 2 2 4 p 1 p B = 3 6 em p 3 p 1 em p p 2....... (p 1) (p 1) em p 2(p 1) em p 2 em p p 2 Lemma 6. Ethe the ght kenel of B s empty, o the fst (p 1)/2 columns of B ae lnealy dependent. Poof. If the ght kenel of B s not empty, then thee exsts (p+1)/2 nteges c 1,..., c (p+1)/2 not dentcally zeo, such that c 1 B,1 + c 2 B,2 +... + c (p+1)/2 B,(p+1)/2 = 0. (6) Apply ths lnea combnaton smultaneously to the fst two ows of B to get c 1 + 2c 2 +... + c (p 1)/2 (p 1)/2 = c (p+1)/2 p (7) 2c 1 + 4c 2 +... + c (p 1)/2 (p 1) = c (p+1)/2 p (8) But (7) mples ethe a contadcton n (8): the ght kenel of B s empty, o c (p+1)/2 = 0 and the fst (p 1)/2 columns of B ae lnealy dependent. Lemma 7. (p + 1)/2 ank(r). Poof. Usng Lemma 6, povng a lowe bound on the ank of R can be educed to showng that the fst (p 1)/2 columns of B ae lnealy ndependent. We use nducton. Consde the sequence of matces B (k) fomed by the fst k columns of B, whee 2 k (p 1)/2. The base case of nducton, B (2), has ank 2 whch s staghtfowad to vefy. Fo the nductve case, we assume B (k 1) has ank k 1, and use Lemma 9 to deduce that B (k) has ank k. The followng lemma s needed befoe povng Lemma 9. Lemma 8. Fo all j 1, (3j em p) 3j = pq fo some ntege q 0.
Ranks of quotents, emandes and p-adc dgts of matces 6 Poof. Wte 3j as 3j = qp + whee = 3j em p and q = 3j quo p. Then 3j = qp. Lemma 9. Let B (k) be the (p 1) k ntege matx n poof of Lemma 7. Ethe B (k) has column ank k, o B (k 1) s ank defcent. Poof. If the ght kenel of B (k) was not empty, then thee exsts nteges c 1,..., c k not dentcally zeo, such that 1 2 k 2 4 2k c 1 3 6 em p (3k) em p. = 0.. (9)... c k 0 We then pefom the followng ow opeatons on the left-hand sde of (9): eplace (ow 3) by (ow 3) 3 (ow 1), then dvde ow 3 by p. Fom Lemma 8, we have that ow 3 s now [ 0 0 1 1 2 q ], fo some q (n fact, q = (3k) quo p). We then pefom the followng column opeatons: let l denote the column ndex whee the fst 1 appeas n ow 3 (l s guaanteed to be geate than o equal 1 snce fo all p > 3, k (p 1)/2, we have 3k > p.) Pvot on enty l n ow 3 and elmnate all entes of ow 3 wth ndces between l + 1 and k 1. Subtact q 1 multples of column l fom column k. Then pvot on enty k of ow 3 and subtact column k fom column l. Effectvely, ths sequence of opeatons tansfoms ow 3 nto: [ 0 0 1 ]. The ght-hand sde of (9) s zeo, and hence not effected by the afoementoned elementay opeatons. Fnally, the tansfomed ow 3 mples ethe that c k s zeo, o the exstence of c 1,..., c k s contadctoy. Ths poves the statement of the lemma. We ae now eady to genealze Lemma 4 and pove Lemma 3. Poof of Lemma 3. Fo the column vecto u Z n 1, consde the matx R Z n n = uu T em p, whch s analogous to the matx R of Lemma 4.
Ranks of quotents, emandes and p-adc dgts of matces 7 The mage of u em p has entes fom the nteval [0, p 1]. If n > p then, by the pgeonhole pncple, the vecto u em p wll contan duplcate (and zeo) entes, whch coespond to duplcate and zeo ows n R. So up to ow/column pemutatons, R contans R as a submatx, and the exta ows/columns ae duplcate and/o zeo. Hence ank( R) = ank(r). 1.3 A Note on Ranks of Latn Squaes It s woth notng that Lemma 4 also mples a esult on the anks of Latn squaes of cetan odes. As befoe, let p be an odd pme, and let R be the (p 1) (p 1) ntege matx whose (, j)th enty s j em p. We show that R s a Latn squae as follows. R s the Cayley multplcaton table of the fnte feld Z/pZ, excludng the element 0. Snce Z/pZ s an ntegal doman, we have j em p j em p wheneve j j (whee, j, j [1, p 1]). So evey ow/column of R has the esdues {1,..., p 1} appeang only once, and R s a Latn squae of ode p 1. R has ank 1 ove Z/pZ and non-tval ank ove Z by Lemma 4 as stated n the followng coollay. Coollay 10. Let p be any odd pme, and let R be any Latn squae of ode p 1 on the symbols {1,..., p 1}. Then the ntege ank of R, taken as a (p 1) (p 1) ntege matx, s (p + 1)/2. 2 p-adc Matces We now swtch the focus to anks of p-adc matces. Ranks n ths secton ae ove the fnte feld wth p elements, wth esdue classes {0, 1,..., p 1}. Fo any pme p and any matx M Z n n wth entes m,j < β, the p-adc expanson of M s M = M [0] + pm [1] +... + p s M [s] whee the entes of each matx M [] ae between [0, p 1], and s log p β. We call M [] the th p-adc matx dgt of M. We extend the supescpt [] notaton to vectos and nteges n the obvous way. We pesent esults concenng the 2-adc matx dgts. Fo odd pmes, we only pesent a conjectue. It s an open queston to study the combnatoal stuctue of the column space of the p-adc matx dgts fo pmes othe than 2. The two anks, ove Z and ove Z/pZ, ae equal unless p s an elementay dvso of the matx.
Ranks of quotents, emandes and p-adc dgts of matces 8 0 0 0 0 0 0 0 1 0 0 1 0 0 1 0 0 1 0 0 0 0 0 1 1 0 1 0 1 0 1 1 0 1 0 0 1 1 0 1 0 1 1 0 0 0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0 1 1 1 1, 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 1 0 1 0 0 1 1 1 0 1 0 0 1 0 0 1 0 1 0 1 0 1 1 0 1 1 0 0 0 1 0 0 1 1 0 0 1 1 0 1 1 1 0 0 0 0 1 0 0 0 1 1 1 0 1 1 1 1 0 1 1 0 0 2 1 1 1 1 0 2 2 1 1 2 0 1 0 1 0 1 2 1 1 0 1 2 1 2 1 2 0 0 1 1 0 1 1 2 0 1 1 2 1 1 2 2 0 1 0 0 1 1 1 0 2 1 1 1 2 2 1 2 0 0 1 0 1 1 0 1 1 2 1 1 2 1 2 2 0 0 0 1 1 0 1 1 1 1 2 1 1 2 2 2 0 1 1 1 0 2 2 2 1 1 1 3 2 2 2 3 0 1 1 0 1 2 1 1 2 2 1 2 3 2 2 3 0 1 0 1 1 1 2 1 2 1 2 2 2 3 2 3 0 0 1 1 1 1 1 2 1 2 2 2 2 2 3 3 0 1 1 1 1 2 2 2 2 2 2 3 3 3 3 4 Fgue 1: An example of A (left) and M = AA T (ght), whee = 4. The ows of A ae pattoned by the numbe of non-zeo entes n each ow. The coespondng blocks n the symmetc matx M ae shown wth bodes. The column pattons of M ae m 0, m 1, m 2, m 3, m 4. And ank p (M [0] ) = ank p (m [0] 1 ) = 4, ank p (M [1] ) = ank p (m [1] 2 ) = 6, ank p (M [2] ) = ank p (m [2] 4 ) = 1. 2.1 Bnay code matces Fx ( p = 2. The goal of ths secton s to show that fo all 1, ank p (M [] ) = ) 2 whee M = AA T fo some specally constucted A, whch we call bnay code matx. We wll genealze the constucton of M n a subsequent secton. Fo now, A s constucted as follows. Stat wth the 2 matx whose, j enty s the jth bt n the bnay expanson of. Then apply ow pemutatons to A such that the fst ( 0) ows have have exactly 0 non-zeo entes, followed by ( 1) ows whch have exactly 1 non-zeo entes, followed by ( 2) ows whch have exactly 2 non-zeo entes and so on. See Fgue 2.1 fo an example whee = 4.
Ranks of quotents, emandes and p-adc dgts of matces 9 The lth column of M s gven by: M,l = a 1,l A,1 +... + a,l A, = j J l A,j, (10) whee J l {1, 2,..., } and the second equalty holds because a,l {0, 1}. We call J l the summng ndex set of M,l. Let m k denote the 2 ( ) k submatx of M, whch ncludes all columns of the fom: M,l = j J l A,j whee J l {1, 2,..., } and J l = k. Then the columns of M can be pattoned nto: M = [ m 0 m 1 m 2... m 2 m 2 +1... m ]. (11) The next lemma shows that [ ] M [] = 0 0... 0 m [] m []... m [] 2 2 +1. (12) Lemma 11. If k < 2, then m [] k = 0 fo all 1. Poof. Columns of m k ae gven by j J A,j whee J = k. The entes of A ae ethe 0 o 1. So the lagest enty n m k s 1 +... + 1 = k. The esult follows by appealng to the bnay expanson of k. We expect ank p (m [] ) ( ) [] 2 2 snce m s a matx of dmenson 2 ( ) 2 2. The next lemma shows that the ank s, n fact, equal to ths uppe bound. Lemma 12. ank p (m [] = ( fo all 1. Poof. Let c 1,..., c ( ) be the column ndces of m 2 n M. Let S(m 2) be the ( 2 ) ( 2 ) 2 submatx of m2 fomed by the ows c 1,..., c ( 2), and S(A) be the ( ) 2 submatx of A fomed by the ows c1,..., c ( 2). Rows of S(A) have exactly 2 non-zeo entes because of the constucton of A. If we teat A and M as block matces then S(m 2 ) = S(A)S(A) T s the 2 th dagonal block of M (See Fgue 2.1). The entes n ow ρ of S(m 2 ) ae gven by lnea combnatons of the entes n ow ρ of S(A). The summng ndex sets J j, whee J j = 2, ae exactly the locatons of the non-zeo entes of ows of S(A), whch ae all dffeent by constucton. Hence thee s only one enty n ow ρ of S(m 2 ) whose summng set matches the locatons of the non-zeo entes n ow ρ of
Ranks of quotents, emandes and p-adc dgts of matces 10 S(A). The value of ths enty s 1 + 1 +... + 1 = 2. The othe entes have values less than 2. Now appeal to the bnay expanson of 2 to get that S(m [] s an dentty (sub)matx of m [] 2 whose sze s ( (. Theefoe, m [] 2 has ank (. Next we wll pove that ank p (M [] ) = ( ) 2 by showng that all the columns of m [], 2 +1 m[],..., 2 +2 m[] 2 ae lnealy dependent on those of m[]. 2 Lemma 13. Consde any column m n m 2 +z, whee z 1. Then m [] s a lnea combnaton of columns of m [] 2. Poof. Let J be the summng ndex set of m, whee J = 2 + z. Let I be the set of all subsets of J of sze 2, so I = ( ) 2 +z 2. Fo evey I I, thee s a unque coespondng column c I n m 2 whose summng set s I. We wll show that m [] can be obtaned by addng up c I s. In othe wods, m [] I I c [] I (mod 2). (13) Let A J denote the submatx of A fomed by the columns ndexed by J. Fo any ow ρ of A J, let 2 + k ρ be the numbe of 1 s n that ow, whee 2 k ρ z. Fst, f k ρ < 0, then the coespondng sum of 1 s at ths ow s less than 2. By Lemma 11, we have the coespondng entes n both m [] 2 and m [] ae zeos and (13) tvally holds. On the othe hand, f 0 k 2 +z ρ z, then the ρth enty of the ght-hand sde of (13) s 1 + 1 +... + 1 ( 2 +k ρ ) 2 (mod 2) snce I = ( 2 +k ρ. (Recall that the numbe of non-zeo entes n ow ρ s 2 +k ρ athe than 2 +z.) The ρth enty of the left-hand sde of (13) s (2 + k ρ ) quo 2. The (2 + k ρ ) tem coesponds to addng (2 + k ρ ) nonzeo entes, and the quo 2 opeaton coesponds to the th bt of the bnay expanson of m. By Lemma 15 (below), we have (2 + k ρ ) quo 2 ( 2 +k ρ ) 2 (mod 2), and (13) holds. The poof of the next (auxlay) lemma uses a theoem due to Kumme [2]. Ths s tue n the example of Fgue 2.1 wthout any eodeng, because we constucted the ow blocks of A such that the bnay expanson of comes afte the bnay expanson of j wheneve > j. Wthout such odeng, the dentty block asseton holds up to ow and column pemutatons.
Ranks of quotents, emandes and p-adc dgts of matces 11 Fact 14 (Kumme s Theoem). The exact powe of p dvdng ( ) a+b a s equal to the numbe of caes when pefomng the addton of (a + b) wtten n base p. A coollay of Kumme s theoem s that ( ) a+b a s odd (esp. even) f addng (a + b) wtten n bnay expanson geneates no (esp. some) caes. Lemma 15. (2 + k) quo 2 ( 2 +k (mod 2). Poof. We wll show that (2 + k) quo 2 and ( ) 2 +k 2 have the same paty. Wte k = Q2 + R fo a quotent Q 0 and a emande 0 R < 2. Thee ae two cases fo Q. If Q s even, then the th bt of k s 0 and hence no caes ae geneated when addng k and 2 n base 2. So by Kumme s Theoem, ( ) ( 2 +k 2 s odd and 2 ) +k 2 1 (mod 2). If Q s odd, then the th bt of k s 1 and the numbe of caes geneated when addng 2 + k n base 2 s at least 1. So by Kumme s theoem ( ) ( 2 +k 2 s even and 2 ) +k 2 0 (mod 2). We have shown that ( 2 +k and Q have opposte pates. Now, substtute k = Q2 + R to get (2 + k) quo 2 = Q + 1. Hence, modulo 2, (2 + k) quo 2 also have an opposte paty to that of Q. Ths concludes ou poof. 2.2 Non-symmetc Matces So fa we have shown that ank p (M [] ) = ank p (m [] = (, whee M = AA T fo some specally constucted A. We now put the esults togethe nto a moe geneal theoem. Theoem 16. Assume U, S, V Z n n, such that U, V have entes fom {0, 1}, det U det V 0 (mod 2), S = dag(1,..., 1, 0,..., 0), ank p (S) =, and n 2. If M = USV Z n n, then ank p (M [] ) = ( fo all 1. Poof. Snce S = SS, we have M = USV = USSV = LR, whee L = US Z n, and R = SV Z n. Let A Z 2 be the bnay code matx of the dgts {0,..., 2 1}. Consde the matces L = A, R = A T and M = L R. If we stat wth L (esp. R) and augment t wth the appopate (n 2 ) addtonal ows (esp. columns), and apply the appopate ow and column pemutatons, then we could tansfom L nto L (esp. R nto R), and n effect, tansfom M nto M. Ou goal s to show that the ank aguments of the pevous lemmas hold unde the afoementoned opeatons..e. the coeffcent of 2 n the bnay expanson of k.
Ranks of quotents, emandes and p-adc dgts of matces 12 We fst note that ow and column pemutatons peseve anks. Also, by a smple enumeaton agument ove the bnay tuples of sze, and by the gven fact that n 2, we can conclude that any addtonal ows (esp. columns) augmented to L (esp. R) wll be lnealy dependent. In fact, any such ows (esp. columns) wll be duplcates of exstng ows (esp. columns). Now, consde addng exta columns to R. The esultng exta columns n M ae duplcates of exstng columns and hence the anks n Lemma 12 ae not affected. Fnally, addng exta ows to L does not change the cadnalty of the summng ndex sets n (10). The est of the esults ae staghtfowad to vefy. 2.3 Odd Pmes Fo p = 2, the non-zeo pattens of the bnay code matx A concdes wth the summng ndces n (10). Ths s not tue fo odd pmes, whee the lnea combnatons can have coeffcents othe than 0 and 1. Thus t s an open queston to devse constucton a smla to bnay code matces, whch exposes the combnatoal stuctue of the column space of M = AA T. Howeve, we pesent the followng conjectue towads undestandng the p- adc anks fo odd pmes. Conjectue 17. Assume p = 2k + 1 s an odd pme, U, S, V Z n n such that U, V have entes fom [0, p 1], det U det V 0 (mod p), S s a 0, 1 dagonal matx and ank p (S) =. Let M = USV = M [0] + M [1] p + whee M [] (Z/pZ) n n. It s conjectued that ank p (M [1] ) k =0 ( ) + 2 + 2 + 1 ( ) + 2k 1 2 (14) 2k Futhemoe, n the genec case whee the entes of U, V ae unfomly chosen at andom fom [0, p 1], and n s abtaly lage, the anks ae equal to the stated bound. Ths conjectue fst appeaed n [1]. It shows that a poduct of matces wth small entes and small ank can stll have vey lage ank, but not full, p-adc expanson. In othe wods, the caes fom the poduct U SV wll mpact many dgts n the expanded poduct.
Ranks of quotents, emandes and p-adc dgts of matces 13 Acknowledgment The authos would lke to thank Andew Anold, Kevn Hae, Davd McKnnon, Jason Peasgood, and B. Davd Saundes. Refeences [1] M. Elshekh, M. Gesbecht, A. Novocn, and B. D. Saundes. Fast computaton of Smth foms of spase matces ove local ngs. In Poceedngs of the 37th Intenatonal Symposum on Symbolc and Algebac Computaton, ISSAC 12, pages 146 153, New Yok, NY, USA, 2012. ACM. [2] E. E. Kumme. Übe de Egänzungssätze zu den allgemenen Recpoctätsgesetzen. Jounal f u de ene und angewandte Mathematk, 44:93 146, 1851. [3] M. Newman. Integal Matces. Academc Pess, New Yok, NY, USA, 1972.